What is the effect of motor loss on an air cooler? 1

[Guylander]

We develop dynamic simulations, and need the opinion of someone with practical experience designing or operating air coolers.

We need to know (ball-park) how much the duty of an air cooler is reduced when the fan power is lost, leaving only natural convection (e.g. Q/Qdesign = 0.5, or 0.25, etc). Of course the easy answer is "it depends on everything" ( design, fin size, process/ambient temperatures, etc.), so any numerical suggestions based on actual experience would be most welcome. Thanks!

 

[speco]

Guylander,

As you know, it really depends on the process temperatures, ambient air temperature, and other factors such as elevation. Of course there is the configuration of the finned tube bundle to consider, too.

However, if you want to get in the ballpark, here's a quick method you might try. The natural convection air flow depends on the natural bouyancy of the heated air to create its own draft, so the numbers can vary a bit. Normally the airside mass velocity is in the range of 5000-6000 pounds per hour per square foot of free area through the bundle. The free area, is typically about half the total face area, but that really depends on the tube pitch, fin OD, fin/inch and fin thickness. If you already know the free area, that's better.

The natural convection mass velocity is usually in the range of 300 to 350 pounds/hr per square foot of free area. You can back-check this to a certain extent by calculating the air temperature rise, to make sure it's not above the inlet process temperature. The U value is generally in the range of 1.0 to 1.5 with this low air velocity.

Hope that helps.

Regards,

Speco (

http://www.stoneprocess.com)

 

[Guylander]

Thanks Speco - sounds like we can use the different flows (e.g. 300, 5000) and find the duty expressions for each to get a rough idea of the effect on Q.

 

[speco]

Guylander,

That's right. If you take a guess at the new heat load, and have an assumed air flow, you can calculate an air temperature rise. From that, you can determine an LMTD. The surface is known. So you can check your assumed heat load and compare it with U*A*LMTD. After a few iterations, you can pretty much zoom in on a heat load where eveything is in balance, at least within reason, and the validity of your assumptions.

Speco

 

[Guylander]

I figured we'd assume a constant process outlet T (valid for single component condenser under pressure control) and an outlet air temperature equal to the process outlet (probably not far off) to compute the duty for the natural draft case.

But the above might not be needed - knowing the typical effect on the air flow is a tremendous help, as the simulated exchanger is designed based on normal conditions, so the simulation can provide the duty and outlet air temp as a function of the air flow.

Thanks again!

 

[CJKruger]

Guylander, for relief load calcs we often assume 20% of normal duty on loss of motor (power failure).

Cilliers

http://www.korf.co.uk

 

[Guylander]

Thanks CJKruger - we've had some plant operators give us similar numbers - between 10%(min) and 30%(max), so 20% seems in line.

 

[786392]

In all fairness existing/prevalent weather conditions will be critical in this connection;e.g. wind speed& the temperature range plus pattern,I believe.

Best Regards
Qalander(Chem)